Download Polypeptides and Proteins

Polypeptides and Proteins
These molecules are composed, at least in part, of chains
of amino acids. Each amino acid is joined to the next one
through an amide or peptide bond from the carbonyl
carbon of one amino acid “residue” to the α-amino group
of the next. At one end of the chain there will be a free or
protonated amino group: the N-terminus. At the other
end there will be a free carboxyl or carboxylate group: the
C-terminus. By convention the N-terminus is drawn at the
left end, as shown below in a generic hexapeptide (6
amino acid residues).
H
H3N C C
R O
N-terminus
H H
H H
H H
H H
H H
N C
C N C
C N C
C N C
C N C
R O
R O
R O
R O
amino acid
residue
planar sections
C O
R O
C-terminus
The peptide linkage is planar --H
C C
O
H
N
C
C
C N
O
C
Note that the backbone chain of amino acid residues
consists of a series of flat “plates” that “join” at the
α-carbons of the amino acid residues. It is the tetrahedral
α-carbons that allow the chain to bend and have some
flexibility. The plates are flat because the amide (or
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peptide) C-N bond is planar owing to its double bond
character as shown by resonance in the figure above.
Because peptide molecules are large a shorthand method
is used to indicate their structures, eg, bradykinin —
H
H
H3N C
C
O
N
C
H
C N
C
O
H H
C
N C
O
H H
C
N C
H O
C
CH2 O
NH
C
HN
NH2
H H
N
C
H
C
N C
CH2 O
H H
C
O
N C
H H
C N
C
C O
CH2 O
O
OH
NH
C
HN
NH2
As you can see, it is simpler to use the three letter
abbreviations for the amino acid residues than it is to
draw the usual structure:
Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg.
2
Note that the N-terminus is at the left end and the
C-terminus is at the right end; it is critical that we hew to
this convention when writing structures using the three
letter abbreviations —
Arg-Phe-Pro-Ser-Phe-Gly-Pro-Pro-Arg would be a
different molecule.
Determination of a polypeptide structure —
1.
Which amino acid residues are present?
2.
How many of each?
3.
What is their sequence?
(1) may be determined by complete acid hydrolysis and
qualitative analysis of the hydrolysate.
(2) may be determined by complete acid hydrolysis,
quantitative analysis of the hydrolysate, and the
determination of the MW of the polypeptide.
Consider a polypeptide, MW = 637. Complete hydrolysis
of 637 mg (1.00 mmole) gives: Asp, 133 mg (1.00 mmole);
Phe, 165 mg (1.00 mmole); Val, 117 mg (1.00 mmole);
Glu, 294 mg (2.00 mmole). Therefore, polypeptide = AspGlu2-Phe-Val, not in sequence.
The sequence (3) may be determined by partial
hydrolysis. Suppose the following units (among others)
are found after partial hydrolysis of the above
polypeptide: Asp-Glu, Val-Asp, Phe-Val, Glu-Glu.
We can extrapolate to the overall sequence as follows:
3
Asp-Glu
Val-Asp
Phe-Val
Glu-Glu
Phe-Val-Asp-Glu-Glu
Therefore, the sequence is Phe-Val-Asp-Glu-Glu.
Another --- real life --- example would be the
determination of the sequence of amino acid
residues in the B-chain of insulin. [Insulin
consists of two polypeptide chains, A and B,
held together by two disulfide bridges, C-S-S-C.
A disulfide bridge results from oxidative coupling
of the S-H groups of two cysteine residues.]
S S
S S
First the disulfide bridges would be cleaved
A
reductively, regenerating cysteine SH groups,
B
and separating the two chains. Then the chain
of interest, B in this case, would be partially hydrolyzed.
The result obtained when this was done follows:
Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu
Ser-His-Leu-Val
Leu-Val-Glu-Ala
Val-Glu-Ala-Leu
Ala-Leu-Tyr
Tyr-Leu-Val-Cys
Val-Cys-Gly-Glu-Arg-Gly-Phe
Gly-Phe-Phe-Tyr-Thr-Pro-Lys
Tyr-Thr-Pro-Lys-Ala
_______________________________________________________________________________________________________________________
Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg-Gly-Phe-Phe-Tyr-Thr-Pro-Lys-Ala
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But how is the sequence in the fragments determined?
Answer: Terminal residue analysis.
Some reagents will react specifically with one terminus or
the other, thus enabling one to determine the nature of
the terminal residue. In some cases, the terminal residue
can be cleaved off and the remainder of the polypeptide
remains (more or less) intact. The remaining polypeptide
can be subjected to terminal residue analysis, etc.
The Edman Degradation —
base
H2NCHCNHCHC
+
Ph N C S
phenyl isothiocyanate
R O R'
polypeptide
H
H
H2O
Ph N C NCHC NHCHC
HCl
R O
R' O
S
N-terminal tagged polypeptide
S
H
C
Ph N
N H
NHCHC
+
C C H
R' O
O
R
degraded peptide
One of 20 possible phenylthiohydantoins,
identified via chromatography, thereby
identifying the terminal a a residue.
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Polypeptides : Smaller than proteins.
MW < ~7-10,000 or < ~50 amino acid residues.
Polypeptides are usually flexible: they usually do not
have the secondary structure characteristic of proteins.
(Nor do they have tertiary or quarternary structure.)
Examples:
bradykinin: helps regulate blood pressure; pain.
methionine enkephalin: Tyr-Gly-Gly-Phe-Met,
morphine like analgesic.
oxytocin:
S
S
cys-tyr-ileu-gln-asn-cys-pro-leu-gly-NH2
causes uterine
contraction and
lactation;
not species specific—
labor induced via chicken oxytocin.
vasopressin (note the structural similarity to oxytocin):
S
S
cys-tyr-phe-gln-asn-cys-pro-arg-gly-NH2
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antidiuretic.
Proteins —
Classification by Gross Chemical Structure
Simple: consist only of amino acid residues.
Conjugated: consist of polypeptide portion(s) and nonpolypeptide portion(s) called prosthetic group(s), eg
lipoproteins, glycoproteins, nucleoproteins,
metalloproteins.
Classification by Gross Morphology
Fibrous: insoluble in water, resist hydrolysis, eg,
sericine (silk fibroin), hard keratin (hair, nails, claws), soft
keratin (skin), myosin (muscle), collagen (connective
tissue, bones).
Globular: soluble or colloidal in water, eg, hemoglobin,
enzymes, antibodies, some hormones.
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Protein Structure
Primary: sequence of amino acid residues.
Secondary: way in which
segments of polypeptide
backbone are oriented in
space; depends on planarity of
peptide bond and, often, Hbonding. The only places in
the protein chain where there
is torsional flexibility is at the
α-carbons: the angles Φ and Ψ
can change causing one plane
to rotate relative to the other.
One could imagine that all of
the planar “plates” in the
protein backbone lie in the
same plane. This would
produce a flat strand. By
snaking back and forth
other parts of the strand
could lie down alongside
the first in parallel (or
antiparallel) fashion
creating a flat sheet. Or
other protein molecules
could lie down parallel (or
antiparallel) to the first.
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This flat sheet would allow hydrogen bonding between
adjacent strands of the protein (favorable for formation of
the sheet). However, the R groups sterically interfere with
each other in this setup. This hypothetical structure has
not been observed in nature.
Antiparallel β-pleated sheet. Black = C, white = H, blue = N, Red = O,
Yellow = R (side chain).
The steric interference between the R groups in the flat
sheet can be reduced if the sheet becomes “pleated”. In
this arrangement the angle between adjacent plates is
not zero. Hydrogen bonding is still possible and now the
R groups can splay away from each other to some extent.
This setup works if the R groups are small. This is the
case for silk fibroin which does form the pleated sheet
structure.
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Another geometry which is
possible is the helix. This
arrangement allows
hydrogen bonding in
directions roughly parallel
to the helical axis (dotted
bonds) and the R groups
are oriented outward from
the helix, away from
each other. This happens
in nature. Several varieties
of helix are known; the
most common form is
known as the α-helix. It is
a right-handed helix (if it
were a wood screw, turning
it clockwise would cause it
to be driven into the wood).
[The mirror image of a
right-handed helix is a left
handed one. The mirror
images of the L-amino
acids in the right-handed
helix would be D-amino
acids. The chirality of the
amino acid residues
determines the chirality of
the helix!]
α-Helix. Black = C, white = H,
blue = N, red = O, purple = R.
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Tertiary: way in which protein as a whole is twisted in
three dimensions; may depend on disulfide bonds, Hbonds, electrostatic attractions, hydrophobic interactions.
The principal factors affecting protein tertiary structure.
Quaternary: aggregate structures held together by
nonbonded interactions or occasionally disulfide links.
Let's look at some proteins!
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The Primary Structure of Bovine Insulin —
Insulin is a protein hormone produced in the pancreas. It
is essential for the regulation of carbohydrate metabolism.
Amino Acid Residues
Species
A-8
A-9
A-10
B-30
Cow
Ala
Ser
Val
Ala
Sheep
Ala
Gly
Val
Ala
Horse
Thr
Gly
Ileu
Ala
Human
Thr
Ser
Ileu
Thr
Pig
Thr
Ser
Ileu
Ala
Whale
Thr
Ser
Ileu
Ala
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I have never seen a discussion of the secondary or
tertiary structure of insulin. It is a small protein so its
secondary structure may be somewhat flexible, although
one would expect it to have conformations of lower and
higher energy with one or several of lower energy
predominating. With regard to tertiary structure the
disulfide links would certainly be important.
Collagen —
Collagen is found in all multicellular
animals and is the most abundant
protein of vertebrates. It is a fibrous
protein, the fibers of which have
great tensile strength. It is found in
the connective tissue of teeth,
bones, ligaments, tendons and
cartilage.
Primary structure —
Collagen is rich in glycine (R=H).
This enables the polypeptide strands
to fit tightly to other strands without
steric difficulties. The other major
components are proline and
hydroxyproline.
Secondary structure —
A left-handed helix.
Tertiary structure —
A right-handed helix.
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Quaternary structure —
A cabling of three right-handed helicies. However, this
continues to higher levels of organization as we can see
in the following photomicrographs.
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Hemoglobin —
Hemoglobin is a complex, conjugated, metalloprotein. It
is what makes red blood cells red. It carries oxygen
throughout the body from the lungs.
Each hemoglobin protein consists of four units, two
α-units and two β-units. This is its quaternary structure.
Each unit consists of a globin and a heme. The globin is
a chain of amino acid
O
Oxygen
residues. The heme is
O
H C CH
CH
a porphyrin containing
CH
one iron atom. The iron
CH
HC
N
N
++
atom carries the oxygen
Fe
Heme
N
N
molecule. The
CH
HC
presence of the nonCH CH COO
OOCCH CH
N
polypeptide heme
makes hemoglobin a
N
H
CH
conjugated protein.
Histidine A A
O
CH
Residue
Each type of unit, α and
HN
C
of Globin
β, has its own particular
shape composed of
various twists and turns. This is its tertiary structure.
2
3
2
3
3
3
-
-
2
2
2
2
2
Within each unit the polypeptide backbone adopts certain
conformations going down the chain – in many places
this is a helix. This is the secondary structure of the
protein.
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Finally, the sequence of the amino acids in the α- (141
residues) and β-globins (146 residues) is the primary
structure.
Oxygenated hemoglobin. The amino acid residues are shown
by number in each globin, starting with #1 at the N-terminus.
This image was drawn looking down the C-2 axis. Ignore the
large grey arrows.
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Over 100 abnormal varieties of hemoglobin have been
reported in humans. The abnormalities are in the globins.
One of these is the replacement of the glutamic acid
residue at position #6 in the β-globins by a valine residue:
Normal:
Val-His-Leu-Thr-Pro-Glu-Lys ---> 146
Abnormal: Val-His-Leu-Thr-Pro-Val-Lys ---> 146
This change causes the hemes associated with the βglobins to be unable to carry oxygen. As a result cells
become oxygen starved.
This alteration also causes the hemoglobin to crystallize
and this, in turn, causes some red blood cells to stiffen
and deform into a crescent shape. These cells can block
blood flow through capillaries. Furthermore, the body
recognizes these cells as abnormal and destroys them,
resulting in anemia. (Life of abnormal cells ~ 16 days;
normal ~ 120 days.) This condition, along with the shape
of the cells gives rise to the name of this genetic disease:
sickle cell anemia.
17
People who
have inherited
two sickle cell
genes (one
from mother,
one from
father) will
have the
disease.
Those who
have inherited
one normal
and one
defective gene
will have some
abnormal
hemoglobin,
but usually
live normal
lives; they are
said to have
the trait, and
can pass the
gene on to
their children.
18
q s
N,s
N,s
y y y y
N,N
N,s
N,s
s,s
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N=normal gene
s=defective gene
N,N=completely
normal
N,s=sickle trait
s,s=sickle cell
disease
Why hasn't this gene disappeared from the gene pool?
Wouldn't it be removed by (Darwinian) natural selection?
Who gets the s gene?
People who live, or whose ancestors lived, in areas where
malaria is, or was, prevalent.
Malaria?
The malarial parasite spends part of its life cycle in red
blood cells. People who are N,s are somewhat resistant
to malaria. Thus, the presence of malaria confers some
advantage to the s gene. This gene did not survive longterm in populations where malaria is not endemic. The
map below shows the relationship between malaria and
the presence of the sickle gene.
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Map of Misery
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